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Integral in the form 1/sqrt

  1. Jun 21, 2007 #1
    integral in the form dx/sqrt(x^2 + a^2)

    1. The problem statement, all variables and given/known data
    # Hi, this is my first post!
    # I have to solve an integral in my electrodynamics study for evaluation of an electric potential. They are in the form [tex]V(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int \rho(\mathbf{r'})/\abs(\mathbf{r}-\mathbf{r'}) d\mathbf{r'}[/tex] In order to find the potential on the top of a conical surface I have to solve an integral in the following form:

    2. Relevant equations
    [tex]\int \frac{dx}{\sqrt{x^2+a^2}}[/tex]

    3. The attempt at a solution
    I solved another problem of the kind
    [tex]\int \frac{x dx}{\sqrt{x^2+a^2}}[/tex]
    but stucked with this one.
    Can you give a hint, or help in order to solve this integral.

    Mathematica gives [tex]\log(x+\sqrt{x^2+a^2})[/tex]
    Last edited: Jun 21, 2007
  2. jcsd
  3. Jun 21, 2007 #2


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    Okay, the attempt at a solution?

    Do you own a handbook on mathematical identites, have you search the internet and/or your books on one-dimensional calculus?
  4. Jun 21, 2007 #3


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    Have you tried any trigonometric or hyperbolic function substutions? Might work out to be arcosh or arsinh of some not so pretty thing, my guess.

    Let's see, int(1/sqrt(x^2+a^2))dx = arsinh (x/a) , a>0
    Last edited: Jun 21, 2007
  5. Jun 21, 2007 #4


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    yes that is correct; [tex]\ln(x+\sqrt{x^2+a^2})[/tex]
    it is.
  6. Jun 21, 2007 #5
    # Thanks for the reply!
    # Here is my attempt. I found this substitution in my Calculus Book and thought it may be useful.
    [tex]x\equiv a\tan(\theta)[/tex]
    so the denominator becomes:
    [tex]a\sqrt{\tan^2(\theta)+1}[/tex] and integral becomes [tex]\frac{\cos{\theta}dx}{a}[/tex]
    and at last the integral after substitution is:
    [tex]\int \frac{d\theta}{\cos{\theta}}[/tex]
    Last edited: Jun 21, 2007
  7. Jun 21, 2007 #6


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    Okay, now I am confused what you want to calculate.

    [tex]V(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int .. [/tex]


    [tex]\int \frac{dx}{\sqrt{x^2+a^2}}[/tex]


    What is the form of the charge distribution?
  8. Jun 21, 2007 #7
    # Interesting, Maxima gives this as an answer. Mathematica and Maxima seems to give different answers but, then, may be this expressions are equal.
  9. Jun 21, 2007 #8
    # Second one. First one is the general formula for vector potentials. But I reduced my problem to the second integral.
  10. Jun 21, 2007 #9


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    Okay, then:

    [tex]\int \frac{dx}{\sqrt{x^2+a^2}} = \ln(x+\sqrt{x^2+a^2})[/tex]
  11. Jun 21, 2007 #10


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    Yes, whenever you see something like x2 + a2, the first thing you'd like to try would be the substitution: x = atan(t).

    After arriving at:
    [tex]\int \frac{d (\theta)}{\cos \theta}[/tex], we notice that the power of the cosine function is 1, ie, odd, so, we'll use the u-substitution: [tex]u = \sin \theta[/tex], like this:
    [tex]\int \frac{d (\theta)}{\cos \theta} = \int \frac{\cos \theta d (\theta)}{\cos ^ 2 \theta} = \int \frac{\cos \theta d (\theta)}{1 - \sin ^ 2 \theta} = \int \frac{du}{1 - u ^ 2}[/tex].
    Hopefully, you can go from here, right?


    Or, a different way to tackle this integral is to use Euler Substitution.
    Let [tex]t = \sqrt{x ^ 2 + a ^ 2} + x[/tex]
    Take the differential of both sides yields:
    [tex]dt = \frac{x}{\sqrt{x ^ 2 + a ^ 2}} dx + dx = \left(\frac{x}{\sqrt{x ^ 2 + a ^ 2}} + 1 \right) dx = \frac{x + \sqrt{x ^ 2 + a ^ 2}}{\sqrt{x ^ 2 + a ^ 2}} = \frac{t}{\sqrt{x ^ 2 + a ^ 2}} dx[/tex]

    Isolate x's, and t's to one side, we have:
    [tex]\Rightarrow \frac{dt}{t} = \frac{dx}{\sqrt{x ^ 2 + a ^ 2}}[/tex]
    Now, integrate both sides gives:
    [tex]\int \frac{dx}{\sqrt{x ^ 2 + a ^ 2}} = \int \frac{dt}{t} = \ln|t| + C = \ln|x + \sqrt{x ^ 2 + a ^ 2}| + C[/tex]


    Or, to make it more convenient, you can try to learn this integral by heart, it's pretty common.

    [tex]\int \frac{dx}{\sqrt{x^2+a^2}} = \ln(x+\sqrt{x^2+a^2}) + C[/tex]
    Last edited: Jun 21, 2007
  12. Jun 21, 2007 #11

    George Jones

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    y = arcsinh(x/a)

    sinhy = x/a

    (e^y - e^-y)/2 = x/a

    Multiply by e^y and rearrange.

    e^(2y) - 2x/a e^y - 1 = 0

    Use the formula for the roots of a quadratic and simplify.

    e^y = (x +/- sqrt(x^2 + a^2))/a

    y = ln(x + sqrt(x^2 + a^2) - lna

    So, up to a constant of integration, the two answers are the same.
  13. Jun 21, 2007 #12
    # Yes, it is the answers. But everywhere what I found is the answer. But I want to learn the intermediate steps too. o:) Otherway I'll feel like I'm cheating in the exam if I memorize the table without learning its proof.
    # Here there are two answers.

    # I tried the substitution x=a*tan(u) and changed the integral to du / cos(u) but stopped there.
  14. Jun 21, 2007 #13


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    hehe yeah, but is it an exam on integral calculus or electrodynamics?..
    Are you not allowed to use collection of formulas and so on?
  15. Jun 21, 2007 #14


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    let x = a sinh (psi)
    dx = a cosh (psi) d(psi)

    then int(1/sqrt(x^2+a^2))dx = int (acosh(psi)/ sqrt((a^2)(sinh(psi))^2 + a^2)) d(psi)

    = int (a cosh (psi)/ a cosh (psi) ) d(psi)
    = int (1) d(psi)
    = psi+C

    but, x = a sinh (psi)
    sihn (psi)= (x/a)
    (psi) = arsinh (x/a)

    so, int(1/sqrt(x^2+a^2))dx = psi+C = arsinh (x/a) +C

    my guess is that ln(x+ sqrt (x^2+a^2)) and this differs by a constant.

    sorry I don't know how to type maths symbols, or this would have looked nicer.
  16. Jun 21, 2007 #15


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  17. Jun 21, 2007 #16


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    thanks, malawi_glenn. =)

    but before I learn all that, the ln(u) thing probably came from the fact that sinh(u)+cosh(u)=e^(u).
  18. Jun 21, 2007 #17
    # Thanks for these important advises. du/(1-u^2) seems much simpler. I think I can solve it looking my Calculus book.

    # And many thanks to everybody! Have a nice day!
  19. Jun 21, 2007 #18
    # It's and electrodynamics exam. In general, the exams are openbook or instructors give some important formulas. But they are all physics books and formulas.
    # And, I think, it's a shame for last year physics students, like me, not able to solve basic integrals.:shy:
    # Thanks for your help.
  20. Jun 21, 2007 #19
    Actually this is true for
    [tex]\int \frac{dx}{\sqrt{x^2 \pm a^2}}=\ln(x+\sqrt{x^2 \pm a^2})[/tex]
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